Answer:
Explanation:
Given that, the range covered by the sphere, 
, when released by the robot from the height, 
, with the horizontal speed is 
as shown in the figure.
The initial velocity in the vertical direction is 
.
Let 
be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. will remain constant throughout the projectile motion.
So, if the time of flight is 
, then
Now, from the equation of motion
Where 
is the displacement is the direction of force, 
is the initial velocity, 
is the constant acceleration and is time.
Here, 
and 
(negative sign is for taking the sigh convention positive in 
direction as shown in the figure.)
So, from equation (ii),
Similarly, for the launched height 
, the new time of flight, 
, is
From equation (iii), we have
Now, the spheres may be launched at speed or 
.
Let, the distance covered in the 
direction be 
for and 
for , we have
[from equation (iv)]
[from equation (i)]
(approximately)
This is in the 
points range as given in the figure.
Similarly, 
[from equation (iv)]
[from equation (i)]
(approximately)
This is out of range, so there is no point for .
Hence, students must choose the speed to launch the sphere to get the maximum number of points.
